Be Prepared for the Calculus Exam Mark Howell Gonzaga High School, Washington, D.C. Martha Montgomery Fremont City Schools, Fremont, Ohio Practice exam contributors: Benita Albert Oak Ridge High School, Oak Ridge, Tennessee Thomas Dick Oregon State University Joe Milliet St. Mark's School of Texas, Dallas, Texas Skylight Publishing Andover, Massachusetts
Copyright 5-9 by Skylight Publishing Chapter 1. Annotated Solutions to Past Free-Response Questions This material is provided to you as a supplement to the book Be Prepared for the AP Calculus Exam (ISBN 978--97755-5-). You are not authorized to publish or distribute it in any form without our permission. However, you may print out one copy of this chapter for personal use and for face-to-face teaching for each copy of the Be Prepared book that you own or receive from your school. Skylight Publishing 9 Bartlet Street, Suite 7 Andover, MA 181 web: e-mail: http://www.skylit.com sales@skylit.com support@skylit.com
9 AB AP Calculus Free-Response Solutions and Notes Question AB-1 (a) a( ) v ( ) ( 8) v( 7) v 7.5 = 7.5 = =.. =.1 8 7 miles/minute. 1 1 (b) vt () dtis the total distance in miles Caren traveled on her trip to school from t = to t = 1 minutes. 1 1 1 1 1 v() t dt =. +. +. +. +.5 +. + 1. = 1.8 miles. (c) Caren turned around at t = minutes. That is when her velocity changed from positive to negative. 1 (d) wtdt () = 1.6 miles is the distance from Larry s home to school. 1 () v t dt= 1. miles is the distance from Caren s home to school. Caren lives closer to school. 1. Or miles per minute per minute.. No need to calculate the final answer. Pay attention to the units on the vertical axis.. Start the calculation at t = 5, when Caren left home the second time, after getting her calculus homework.
FREE-RESPONSE SOLUTIONS ~ 9 AB Question AB- (a) R () t dt 1 = 98 people. =. R ( t) ( ) =, R( 1.6) = 85.57, and ( ) (b) R () t 18t 675 t R maximum at t = 1.6 hours. (c) ( ) () 1 = () = ( ) () w w w t dt t R t dt 1 1 = at t = and t = 1.6 hours. R = 1. By the candidate test, R(t) has the = 87.5 hours. (d) Total wait time is w () t dt = ( ) () waits 76.776 98 = hours. t R t dt = 76. On average, a person 1. Use the given function name in your formulas. 1 1 Or, if you are using symbolic antidifferentiation, 18 8 675 16.
FREE-RESPONSE SOLUTIONS ~ 9 AB 5 Question AB- (a) It costs 5 6 xdx = 5 dollars to produce the cable. Mighty receives $1 5 meters = $ for the cable. The profit is $5. meter (b) 6 xdx is the cost difference in dollars between manufactoring a meter long 5 cable and a 5 meter long cable or the additional cost to manufacture an additional 5 meters at the end of a 5 meter cable. 1 P k = 1k k 6 xdx dollars. (c) Profit is ( ) (d) P ( k) = 1 6 k k =. Since P ( k) > for < k < and P ( k) < for k >, profit is a maximum when k =. The maximum profit is P( ) = 1 6 xdx= 16 dollars. 1. No need to evaluate this integral. Question AB- (a) Area= x x x dx x 8 = =. π π = = + =. π π π (b) Volume = sin x dx cos x ( 1 1) (c) Volume = y y dy. 1 1. Not ( ) x x dx. The sections are perpendicular to the y-axis, not the x-axis.
6 FREE-RESPONSE SOLUTIONS ~ 9 AB Question AB-5 (a) f ( ) ( ) f ( ) f 5 = =. 5 ( f x ) dx ( x f ( x) ) ( ) ( ) 1 1 (b) ( ) 5 = 5 = 9 6 5 = 8 (c) Left Riemann sum = 11 + + ( ) + 5 = 18. 1 (d) The tangent line at x = 5 is y ( x 5) Since f ( x) and the tangent line is above the curve. Thus, f ( 7).. 1 = +. On the tangent line, at x = 7, y =. < for all x in [5, 8], the graph of f is concave down on that interval, ( ) 5 The slope of the secant line = =. An equation of the secant line is 8 5 5 1 y = + ( x 5). At x = 7, y = + =. The secant line is under the graph of f, so f ( 7). 1. Calculating the final answer is optional.
FREE-RESPONSE SOLUTIONS ~ 9 AB 7 Question AB-6 (a) The graph of f has points of inflection only at x = and x =, since these are the only values of x where f has local extrema. 1 (b) ( ) ( ) ( ) ( ) f = f + f x dx= 5 8 π = π. ( ) ( ) ( ) x x ( ) ( ) ( ) ( ) f = f + f x dx= / / / / e dx e x e e 5 + 5 = 5 + 15 = 5 + 15 1 15 = 8 15 5 for < x < ln, f is increasing there. Since ( ) f x < for 5 ln < x <, f is decreasing there. Therefore, f has its absolute maximum at 5 x = ln. (c) Since f ( x) 1. That is, f changes from decreasing to increasing and vice-versa at these values of x..
9 BC AP Calculus Free-Response Solutions and Notes Question BC-1 See AB Question 1. Question BC- See AB Question. 9
1 FREE-RESPONSE SOLUTIONS ~ 9 BC Question BC- dy (a) dt = when t =.677. Let B = dy.677. 1 For < t < B, > and for dt dy t > B, <. Therefore, y is maximum at t = B. dt yb ( ) = y() +.6 9.8tdt B 1.61 meters. (b) t. yt ( ) = y() +.6 9.8udu= 11. +.6t.9t y() t = at t = 1.9656 seconds. 1 A = 1.96 seconds. 1 (c) Total distance traveled is A dx dy + dt dt 1.96 meters. dy dy dt.6 9.8A = = = dx 19.1911. 1, dx t= A.8 dt t= A The path makes an angle of arctan ( tanθ ) 1.519 radians with the water's surface. (d) tan ( θ ) 1. Store this value in your calculator.. π dy We need the absolute value because < θ <. Here = tan( π θ) = tan( θ). dx t= A. The work in this problem is easy to check. Plot the motion of the diver in parametric mode and verify your answers.
FREE-RESPONSE SOLUTIONS ~ 9 BC 11 Question BC- dy, 6 dx = =. So, 1 y new = + =. At dy 6 1 1 1 dx = =. So 1 1 y = + =.5. (a) At the point ( 1, ) 1,, (b) T ( x) ( x 1) ( x 1) (c) x ( 6 y) 1. Or: 1 = + + + +. dy =. The constant solution y = 6 does not satisfy the initial condition. dx dy Separating the variables, x dx 6 y =. Antidifferentiating, we have x 1 ln 6 y = + C. Substituting x= 1, y = we get ln = + C 1 x 1 x 1 C = ln. Therefore, ln 6 y = + ln ln 6 y = + ln x 1 ln 6 y = e + x 1 ln y = 6 e +. 1 x ln 6 y = + C 6 y = Ae (where x A=± e C ). Substituting 1 1 1 x x= 1, y = we get = Ae A= e y = 6 e e. Question BC-5 See AB Question 5.
1 FREE-RESPONSE SOLUTIONS ~ 9 BC Question BC-6 (a) ( ) ( ) ( ) 6 n 1 1 1 1 ( x ) x x x e = 1+ ( x 1 ) + + +... + +... 1 6 n! (b) ( x 1) ( x 1) ( x 1) ( x 1) 6 n 1 + + +... + +... 6! n! (c) ( x 1) ( n ) n ( x 1) n ( x ) + 1! 1 lim = lim = < 1 n n n + 1 n! real numbers. for all x. The interval of convergence is all (d) f ( x) ( x ) ( x ) ( )( ) n n x 1 = 1 + 1 +... + +... n! ( )( )( ) n n n x 1 f ( x) = 1+ ( x 1 ) +... + +... n! f is positive for all x, so the graph of f has no points of inflection. 1. Just substitute ( x 1) for x in the given series for. Note that f(1) = 1 is given; just divide each term by x e. ( x 1).
9 AB (Form B) AP Calculus Free-Response Solutions and Notes Question AB-1 (Form B) ( ) ( ) 1 R t u du t (a) () = 6+ + sin( ) 16 1 ( ) = 6+ + sin( ) 6.61877 R u du 16 (b) At () = π ( Rt ()) da dr = π R() t dt da dt dt dr 1 = πr( ) = π R() ( + sin9) dt 16 t= t= 1 (( ) ( ( )) ) 6.611 centimeters. 8.858 cm / year. (c) A ( t) dt = A( ) A( ) = π R( ) R.1 cm. From t = to t = years, the area of a cross section of the tree at the given height changed by.1 cm. 1. Save this value in your calculator. 1
1 FREE-RESPONSE SOLUTIONS ~ 9 AB (FORM B) Question AB- (Form B) 5 (a) 5 f () t dt + 6.95 meters. (b) At time t = hours, the rate of change of the distance from the water s edge to the road was increasing at the rate of 1.7 meters/hour per hour. (c) We want the minimum value of f(t). f ( t) = at t 1 =.6619 and at t =.8. Using the candidate test: f =, f t = 1.98, f t =.7, f 5 =.8. The minimum ( ) ( ) ( ) ( ) 1 occurs at t.8 hours. t = + g p dp, where t is the number of days that sand must be pumped (d) 5 6.95 ( ) to restore the original distance between the road and the edge of the water.
FREE-RESPONSE SOLUTIONS ~ 9 AB (FORM B) 15 Question AB- (Form B) (a) (b) (c) ( ) ( ) f x f No. We can see from the graph that is equal or close to x for f ( x) f ( ) < x <. Therefore, lim >. We can see from the graph that f is x x f ( x) f ( ) decreasing on (, ), so is negative for < x <. 1 Therefore, x f ( x) f ( ) f ( x) f ( ) lim (or doesn t exist). Thus lim does not exist. + x x x x Two. For the average rate of change of f on the interval [a, 6] to be, we need f ( 6) f ( a) = 6 a again between and., or f ( a ) = 1. This happens twice: once between and 1 and ( ) f ( ) f 6 1 1 Yes, a =. On the closed interval [, 6], we have = =. Since f 6 is continuous on [, 6] and differentiable on (, 6), the Mean Value Theorem 1 guarantees the existence of a number c such that < c < 6 and f ( c) =. (d) g ( x) = f ( x) and g ( x) f ( x) have, f ( x) =. Since f is increasing on (, ) and (, 6), we > on these intervals, so g ( x) > and the graph of g is concave up on (, ) and (, 6). 1. Is the graph alone a sufficient justification? We ll know for sure only when the rubric is published. It is probably safer to add the following: Indeed, f ( x) > on (, ), so f is increasing there; f () =, so f ( x) < on (, ), so f is decreasing there.. Or on [, ] and [, 6].
16 FREE-RESPONSE SOLUTIONS ~ 9 AB (FORM B) Question AB- (Form B) (a) Area = / x 16. 1 x x dx= x = (b) Volume = 6 6 8 +. 5 1 x / x x 5/ x x dx= x x + dx= x + = 5 1 x (c) Volume = π ( x ) dx. 1.. =. 8 =. 15
FREE-RESPONSE SOLUTIONS ~ 9 AB (FORM B) 17 Question AB-5 (Form B) ( ) (a) g ( x) = f ( x) e f x f ( 1) g ( 1) = f ( 1) e = e. g( 1) e equation is y e = e ( x 1). 1 (b) Since ( ) =. The tangent line f x e > for all x, the sign of g ( x) is the same as the sign of f ( x). Therefore, g ( x) changes sign from positive to negative only at x = 1, so g has a local maximum only at x = 1. (c) ( ) ( 1) (( ( )) ( )) f g 1 = e f 1 + f 1. (because f ( x) is decreasing around 1 f ( 1 ) e >, ( ) x = ), so ( ) f 1 =, and f ( 1) < g 1 <, (d) From Part (a), g () 1 = e. The average rate of change of g on [1, ] is ( ) ( 1) + g g e = = e. 1 1. Or y 5e e x =.
18 FREE-RESPONSE SOLUTIONS ~ 9 AB (FORM B) Question AB-6 (Form B) (a) a ( 6) ( ) v( ) v 7+ 11 = = 8 8 meters/second. (b) vtdt () is the net change in meters of the position of the particle from t = seconds to t = seconds. The trapezoidal sum approximation is 1 + ( 8) 8 + ( ) + 7 5+ 7+ 8 meters. 1 (c) Yes. The particle must change direction from right to left in the interval [8, ] when its velocity changes from positive to negative. The particle must change direction from left to right in the interval [, ] when its velocity changes from negative to positive. 8 8 (d) The position of the particle at t = 8 ( 8) = ( ) + ( ) = 7+ ( ). x x v t dt v t dt () = v () t > for < t < 8, so v( t ) is increasing and v( t) v() a t 8 interval. Therefore, vtdt> () 8 = 1. = 75 meters. and, ( ) > = on that time x 8 > 7 + > meters.
Question BC-1 (Form B) (a) Area = ( ) f x dx 9 BC (Form B) AP Calculus Free-Response Solutions and Notes 18.8. 1 (b) π πx The volume of the cake V = 1sin dx 56.19 cm. There will be.5 56.19 117.81 grams of chocolate in the cake. (c) Perimeter = + 1+ ( f ( x) ) dx 81.8 cm. 1. Or π x 6 cos 6 1 π =. π Question BC- (Form B) See AB Question. Question BC- (Form B) See AB Question. 19
FREE-RESPONSE SOLUTIONS ~ 9 BC (FORM B) Question BC- (Form B) π (a) The graph of the curve goes through the origin when 1 cosθ = θ =. 1 π / Area = ( 1 cos ) d θ θ. dx x = θ θ = θ θ sinθ cosθsinθ dθ = +. dy y = ( 1 cosθ) sinθ = sinθ cosθ sinθ cosθ sin θ cos θ dθ = +. 1 (b) ( 1 cos ) cos cos cos (c) π At θ =, r = 1, x =, y = 1, and is y = 1 x. dy dy dx dx = dθ dθ = 1 =. Tangent line equation dy y = 1 cosθ sinθ = sinθ sin θ cosθ cos θ dθ =. 1. Or ( ) Question BC-5 (Form B) See AB Question 5.
FREE-RESPONSE SOLUTIONS ~ 9 BC (FORM B) 1 Question BC-6 (Form B) (a) This series is a geometric series with common ratio x + 1. It converges when x + 1 < 1 that is, when < x <. (b) The sum = 1 1 =. 1 + 1 x ( x ) x 1 x g x = dt t ln x 1 t = = (c) ( ) ln 1. Thus, g 1 1 = ln. 1 (d) h( x) x x x n = 1 + + +... + +... 1 1 1 h = f 1 = f = =. 1. = ln.